Add solution #1567

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,372 LeetCode solutions in JavaScript
+# 1,373 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1196,6 +1196,7 @@
 1562|[Find Latest Group of Size M](./solutions/1562-find-latest-group-of-size-m.js)|Medium|
 1563|[Stone Game V](./solutions/1563-stone-game-v.js)|Hard|
 1566|[Detect Pattern of Length M Repeated K or More Times](./solutions/1566-detect-pattern-of-length-m-repeated-k-or-more-times.js)|Easy|
+1567|[Maximum Length of Subarray With Positive Product](./solutions/1567-maximum-length-of-subarray-with-positive-product.js)|Medium|
 1576|[Replace All ?'s to Avoid Consecutive Repeating Characters](./solutions/1576-replace-all-s-to-avoid-consecutive-repeating-characters.js)|Medium|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|

solutions/1567-maximum-length-of-subarray-with-positive-product.js

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+/**
+ * 1567. Maximum Length of Subarray With Positive Product
+ * https://leetcode.com/problems/maximum-length-of-subarray-with-positive-product/
+ * Difficulty: Medium
+ *
+ * Given an array of integers nums, find the maximum length of a subarray where the product of
+ * all its elements is positive.
+ *
+ * A subarray of an array is a consecutive sequence of zero or more values taken out of that array.
+ *
+ * Return the maximum length of a subarray with positive product.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var getMaxLen = function(nums) {
+  let result = 0;
+  let positiveCount = 0;
+  let negativeCount = 0;
+  let firstNegativeIndex = -1;
+
+  for (let i = 0; i < nums.length; i++) {
+    if (nums[i] === 0) {
+      positiveCount = 0;
+      negativeCount = 0;
+      firstNegativeIndex = -1;
+      continue;
+    }
+
+    if (nums[i] > 0) {
+      positiveCount++;
+    } else {
+      negativeCount++;
+      if (firstNegativeIndex === -1) {
+        firstNegativeIndex = i;
+      }
+    }
+
+    if (negativeCount % 2 === 0) {
+      result = Math.max(result, positiveCount + negativeCount);
+    } else {
+      result = Math.max(result, i - firstNegativeIndex);
+    }
+  }
+
+  return result;
+};